AC自动机 HDU 2222 Keywords Search

Problem Description

In the modern time, Search engine came into the life of everybody like Google, Baidu, etc.
Wiskey also wants to bring this feature to his image retrieval system.
Every image have a long description, when users type some keywords to find the image, the system will match the keywords with description of image and show the image which the most keywords be matched.
To simplify the problem, giving you a description of image, and some keywords, you should tell me how many keywords will be match.

 

Input

First line will contain one integer means how many cases will follow by.
Each case will contain two integers N means the number of keywords and N keywords follow. (N <= 10000)
Each keyword will only contains characters 'a'-'z', and the length will be not longer than 50.
The last line is the description, and the length will be not longer than 1000000.

 

Output

Print how many keywords are contained in the description.

 

Sample Input

1 5 she he say shr her yasherhs

 

Sample Output

3

 

 

AC自动机资料：http://www.pin5i.com/showtopic-27128.html

                     http://blog.henix.info/blog/trie-aho-corasick.html

                     http://blog.csdn.net/niushuai666/article/details/7002823

                     http://yzmduncan.iteye.com/blog/1217014

 

模板：

#include <stdio.h>
#include <stdlib.h>

int const N= 500010;
struct Trie{
    int flag;  // 标记是否为某一模式串的结尾 
    int fail;  // 失败指针 
    int next[26];
    void init(){
        flag= 0; fail= -1;
        for( int i= 0; i< 26; ++i ) next[i]= 0; }
}tb[N];

int cnt= 0, que[N], n;
char str[1000010];

void inline insert( char* s ){
    int rt= 0;
    while( *s ){
        int t= *s- 'a';
        if( !tb[rt].next[t] ){
            tb[++cnt].init();
            tb[rt].next[t]= cnt;
        }
        rt= tb[rt].next[t]; s++;
    }
    tb[rt].flag++;
}

void bfs(){
    int head= 0, tail= 0, p, q;
    que[0]= 0;
    while( head<= tail ){
        int now= que[head++];
        for( int t= 0; t< 26; ++t )
        if( tb[now].next[t] ){
            p= tb[now].fail, q= tb[now].next[t];
            while( p!= -1 && !tb[p].next[t] ) p= tb[p].fail;
            if( p== -1 ) tb[q].fail= 0;
            else tb[q].fail= tb[p].next[t];
            que[++tail]= q;
        }
    }
}

void Match( char* s ){
    int ans= 0, rt= 0, t, p;
    while( *s ){
        t= *s- 'a';
        if( tb[rt].next[t] ) rt= tb[rt].next[t];
        else{
            p= tb[rt].fail;
            while( p!= -1 && !tb[p].next[t] ) p= tb[p].fail;
            if( p== -1 ) rt= 0;
            else rt= tb[p].next[t];
        }
        p= rt;
        while( p!= 0 && tb[p].flag ){
            if( tb[p].flag ){
                ans+= tb[p].flag; tb[p].flag= 0; }
            p= tb[p].fail;
        }
        s++;
    }
    printf("%d\n", ans );
}

int main(){
    int test;
    scanf("%d",&test ); 
    while( test-- ){
        scanf("%d\n",&n ); 
        cnt= 0; tb[0].init();
        while( n-- ){
            gets(str);
            insert( str );
        }
        bfs();
        gets(str);
        Match( str );
    }
    
    return 0;
}